Question: Solve for $x$ : $ 3|x + 1| - 4 = -5|x + 1| + 8 $
Answer: Add $ {5|x + 1|} $ to both sides: $ \begin{eqnarray} 3|x + 1| - 4 &=& -5|x + 1| + 8 \\ \\ { + 5|x + 1|} && { + 5|x + 1|} \\ \\ 8|x + 1| - 4 &=& 8 \end{eqnarray} $ Add ${4}$ to both sides: $ \begin{eqnarray} 8|x + 1| - 4 &=& 8 \\ \\ { + 4} &=& { + 4} \\ \\ 8|x + 1| &=& 12 \end{eqnarray} $ Divide both sides by ${8}$ $ \dfrac{8|x + 1|} {{8}} = \dfrac{12} {{8}} $ Simplify: $ |x + 1| = \dfrac{3}{2}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{3}{2} $ or $ x + 1 = \dfrac{3}{2} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{3}{2} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{3}{2} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{3}{2} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $2$ $ x = - \dfrac{3}{2} {- \dfrac{2}{2}} $ $ x = -\dfrac{5}{2} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{3}{2} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{3}{2} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{3}{2} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $2$ $ x = \dfrac{3}{2} {- \dfrac{2}{2}} $ $ x = \dfrac{1}{2} $ Thus, the correct answer is $x = -\dfrac{5}{2} $ or $x = \dfrac{1}{2} $.